Linear Equations in Linear Algebra - Chapter 1 - Counterexample on Unique Echelon Form

Oct 28, 2026

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Solutions-style content addressing a key claim in linear algebra: that every matrix is row equivalent to a unique matrix in echelon form. The text explains a counterexample strategy using a starting matrix A, transforming it by different sequences of row operations into two distinct echelon-form matrices X and Y. It concludes that both X and Y are row equivalent to A, so echelon-form equivalents need not be unique. The material references Chapter 1 and nearby sections/exercises.

LanguageEnglish

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Chapter 1: Linear Equations in Linear Algebra (cont'd)

Section 1.8 - Section 1.9

Section 1.10

Supplementary Exercises

Linear algebra and its applications 6th edition solutions manual pdf The statement that every matrix is row equivalent to a unique matrix in echelon form is false. This can be demonstrated with a counterexample, where two different matrices (X and Y) are shown to be row equivalent to the same matrix in echelon form. To illustrate this, consider the following steps: 1 . Start with a given matrix A. 2. Apply row operations to transform A into matrix X, which is in echelon form. 3. Then, apply another set of row operations to transform A into matrix Y, which is also in echelon form. By doing so, it can be shown that matrices X and Y are both in echelon form and are row equivalent to the original matrix A. This means that not every matrix has a unique equivalent matrix in echelon form. In essence, this shows that the statement "every matrix is row equivalent to a unique matrix in echelon form" is false, as there can be multiple matrices in echelon form that are row equivalent to a given matrix. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 Chapter 1: Linear Equations in Linear Algebra (cont'd 3) Section 1.8 Section 1.9 Exercise 1 Exercise 9 Exercise 17 Exercise 25 Exercise 33 Exercise 1 Exercise 9 Exercise 17 Exercise 25 Exercise 33 Exercise 2 Exercise 10 Exercise 18 Exercise 26 Exercise 34 Exercise 2 Exercise 10 Exercise 18 Exercise 26 Exercise 34 Exercise 3 Exercise 11 Exercise 19 Exercise 27 Exercise 35 Exercise 3 Exercise 11 Exercise 19 Exercise 27 Exercise 35 Exercise 4 Exercise 12 Exercise 20 Exercise 28 Exercise 36 Exercise 4 Exercise 12 Exercise 20 Exercise 28 Exercise 36 Exercise 5 Exercise 13 Exercise 21 Exercise 29 Exercise 37 Exercise 5 Exercise 13 Exercise 21 Exercise 29 Exercise 37 Exercise 6 Exercise 14 Exercise 22 Exercise 30 Exercise 38 Exercise 6 Exercise 14 Exercise 22 Exercise 30 Exercise 38 Exercise 7 Exercise 15 Exercise 23 Exercise 31 Exercise 39 Exercise 7 Exercise 15 Exercise 23 Exercise 31 Exercise 39 Exercise 8 Exercise 16 Exercise 24 Exercise 32 Exercise 40 Exercise 8 Exercise 16 Exercise 24 Exercise 32 Exercise 40 Chapter 1: Linear Equations in Linear Algebra (cont'd 4) Section 1.10 Supp. Exercises Exercise 1 Exercise 6 Exercise 11 Exercise 1 Exercise 6 Exercise 11 Exercise 16 Exercise 21 Exercise 2 Exercise 7 Exercise 12 Exercise 2 Exercise 7 Exercise 12 Exercise 17 Exercise 22 Exercise 3 Exercise 8 Exercise 13 Exercise 3 Exercise 8 Exercise 13 Exercise 18 Exercise 23 Exercise 4 Exercise 9 Exercise 14 Exercise 4 Exercise 9 Exercise 14 Exercise 19 Exercise 24 Exercise 5 Exercise 10 Exercise 5 Exercise 10 Exercise 15 Exercise 20 Exercise 25 Back to Top Top

FAQs

Why is the statement about a unique row-equivalent echelon form false?

The text shows that a single matrix A can be transformed by different row-operation sequences into two different echelon-form matrices X and Y.

How are matrices X and Y obtained in the counterexample?

Starting from A, one applies row operations to reach X in echelon form, then applies another set of row operations to reach Y in echelon form.

What is the key conclusion about echelon-form matrices that are row equivalent to the same matrix?

Multiple echelon-form matrices can be row equivalent to a given matrix, so uniqueness does not hold.

Oct 28, 2026

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